Circles and Squares

Date: 4/3/96 at 17:12:7
From: Anonymous
Subject: Area of Sphere-Cube and Inside Sphere
Good day. This is Jared Martin, a Pre-Algebra student in seventh
grade at West Cary in Cary, NC. My class has been working on the
geometry chapter in our books. I was fooling around with my math
one day and I began to think about 3D and 2D circles, one large
size circle with a square inside the circle (the sides of the
square are equal to the radius of the larger circle but are
actually chords in the circle) and a smaller circle inside the
square, who's radius is equal to one-half of the square's side.
My theory states as follows:
The diameter of the larger circle times pi is divided by two.
This product is then squared and multiplied by pi and the length
of side of the square. The center points of the three objects are
the same.
My question is:
What is the percent of area of the small circle compared to the
large circle's area?
Please answer if the larger circle's diameter is 9m.
Thank you for your time and effort.
Jared

Date: 4/6/96 at 15:48:8
From: Doctor Aaron
Subject: Re: Area of Sphere-Cube and Inside Sphere
Hi Jared,
You've been thinking about some neat stuff, but I'm not sure
exactly what you mean. You specified a square inside of a circle,
where the side of the square is equal in length to the radius of
the circle and the sides of the square are chords in the circle.
The thing is that we can't get all four sides of the square to be
chords if we make the length equal to the radius of the circle.
To answer your question I'm going to have to drop one of these
assumptions, and the more interesting problem is the one where the
sides of the square are chords in the circle. In this case we say
that the square is _inscribed_ in the circle. For ease of
notation I'm going to define a few variables.
Let r = the radius of the small circle.
From this we can uniquely determine:
s = the side of the square
R = the radius of the large circle.
Once we have these, we'll be able to get the areas of all of the
shapes.
Okay. You described the small circle as having radius of length
equal to half of the squares side. Then we say that the small
circle is inscribed in the square because it touches the square
without overlapping it. Another way to put this is that the
circle is tangent to the square at 4 points (the midpoints of the
sides). Well, we know s = 2r, but finding R is a little harder.
Draw a square inscribed in a circle. Then draw the radius that
connects the center of both objects to a corner of the square.
Next draw a radius in the small circle to connect the center to
the midpoint of one of the sides of the square that defines the
corner that you drew the first radius to. Whoa, that was wordy.
You should see a right triangle that composes one eighth of the
square. If you use Pythagoras you can get the radius of the big
circle. If you haven't learned the Pythagorean theorem yet, you
will have to prove it to solve this problem.
The other case you talked about was a sphere inscribed in a cube
inscribed in a sphere. This is also very interesting. You can
use the same techniques to go from the small sphere to the big
sphere except you have to use the Pythagorean theorem twice
because we will get a right triangle that is not situated in the
plane of any of the faces of the cube so we must decompose it.
I hope that these hints are helpful. Keep thinking about math.
-Doctor Aaron, The Math Forum